Exam (elaborations) TEST BANK FOR McMurry's Organic Chemistry 7th Edit

Exam (elaborations) TEST BANK FOR McMurry's Organic Chemistry 7th Edit Van't Hoff and Le Bel proposed that the 4 atoms to which carbon forms bonds sit at the corners of a regular tetrahedron. 4. In a drawing of a tetrahedral carbon, a wedged line represents a bond pointing toward the viewer, a dashed line points behind the plane of the page, and a solid line lies in the plane of the page.. B. Covalent bonds. 1. Atoms bond together because the resulting compound is more stable than the individual atoms. a. Atoms tend to achieve the electron configuration of the nearest noble gas. b. Atoms in groups 1A, 2A and 7A either lose electrons or gain electrons to form ionic compounds. c. Atoms in the middle of the periodic table share electrons by forming covalent bonds. d. The neutral collection of atoms held together by covalent bonds is a molecule. 2. Covalent bonds can be represented two ways. a. In electron-dot structures, bonds are represented as pairs of dots. b. In line-bond structures, bonds are represented as lines drawn between two bonded atoms. 3. The number of covalent bonds formed by an atom depends on the number of electrons it has and on the number it needs to achieve an octet. 4. Valence electrons not used for bonding are called lone-pair (nonbonding) electrons. a. Lone-pair electrons are often represented as dots. C. Valence bond theory (Section 1.5). 1. Covalent bonds are formed by the overlap of two atomic orbitals, each of which contains one electron. The two electrons have opposite spins. 2. Bonds formed by the head-on overlap of two atomic orbitals are cylindrically symmetrical and are called σ bonds. 3. Bond strength is the measure of the amount of energy needed to break a bond. 4. Bond length is the optimum distance between nuclei. 5. Every bond has a characteristic bond length and bond strength. III. Hybridization (Sections 1.6–1.10). A. sp3 Orbitals (Sections 1.6, 1.7). 1. Structure of methane (Section 1.6). a. When carbon forms 4 bonds with hydrogen, one 2s orbital and three 2p orbitals combine to form four equivalent atomic orbitals (sp3 hybrid orbitals). b. These orbitals are tetrahedrally oriented. c. Because these orbitals are unsymmetrical, they can form stronger bonds than unhybridized orbitals can. d. These bonds have a specific geometry and a bond angle of 109.5°. 2. Structure of ethane (Section 1.7). a. Ethane has the same type of hybridization as occurs in methane. b. The C–C bond is formed by overlap of two sp3 orbitals. c. Bond lengths, strengths and angles are very close to those of methane. B. sp2 Orbitals (Section 1.8). 1. If one carbon 2s orbital combines with two carbon 2p orbitals, three hybrid sp2 orbitals are formed, and one p orbital remains unchanged. 2. The three sp2 orbitals lie in a plane at angles of 120°, and the unhybridized p orbital is perpendicular to them. 3. Two different types of bonds form between two carbons. a. A σ bond forms from the overlap of two sp2 orbitals. b. A π bond forms by sideways overlap of two p orbitals. c. This combination is known as a carbon–carbon double bond. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Structure and Bonding 3 4. Ethylene is composed of a carbon–carbon double bond and four σ bonds formed between the remaining four sp2 orbitals of carbon and the 1s orbitals of hydrogen. a. The double bond of ethylene is both shorter and stronger than the C–C bond of ethane. C. sp Orbitals (Section 1.10). 1. If one carbon 2s orbital combines with one carbon 2p orbital, two hybrid sp orbitals are formed, and two p orbitals are unchanged. 2. The two sp orbitals are 180° apart, and the two p orbitals are perpendicular to them and to each other. 3. Two different types of bonds form. a. A σ bond forms from the overlap of two sp orbitals. b. Two π bonds form by sideways overlap of four unhybridized p orbitals. c. This combination is known as a carbon–carbon triple bond. 4. Acetylene is composed of a carbon–carbon triple bond and two σ bonds formed between the remaining two sp orbitals of carbon and the 1s orbitals of hydrogen. a. The triple bond of acetylene is the strongest carbon–carbon bond. D. Hybridization of nitrogen and oxygen (Section 1.10). 1. Covalent bonds between other elements can be described by using hybrid orbitals. 2. Both the nitrogen atom in ammonia and the oxygen atom in water form sp3 hybrid orbitals. a. The lone-pair electrons in these compounds occupy sp3 orbitals. 3. The bond angles between hydrogen and the central atom is often less than 109° because the lone-pair electrons take up more room than the σ bond. 4. Because of their positions in the third row, phosphorus and sulfur can form more than the typical number of covalent bonds. IV. Molecular orbital theory (Section 1.11). A. Molecular orbitals arise from a mathematical combination of atomic orbitals and belong to the entire molecule. 1. Two 1s orbitals can combine in two different ways. a. The additive combination is a bonding MO and is lower in energy than the two hydrogen 1s atomic orbitals. b. The subtractive combination is an antibonding MO and is higher in energy than the two hydrogen 1s atomic orbitals. 2. Two p orbitals in ethylene can combine to form two π MOs. a. The bonding MO has no node; the antibonding MO has one node. 3. A node is a region between nuclei where electrons aren't found. a. If a node occurs between two nuclei, the nuclei repel each other. V. Chemical structures (Section 1.12). A. Drawing chemical structures. 1. Condensed structures don't show C–H bonds and don't show the bonds between CH3, CH2 and CH units. 2. Skeletal structures are simpler still. a. Carbon atoms aren't usually shown. b. Hydrogen atoms bonded to carbon aren't usually shown. c. Other atoms (O, N, Cl, etc.) are shown. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 4 Chapter 1 Solutions to Problems 1.1 (a) To find the ground-state electron configuration of an element, first locate its atomic number. For oxygen, the atomic number is 8; oxygen thus has 8 protons and 8 electrons. Next, assign the electrons to the proper energy levels, starting with the lowest level. Fill each level completely before assigning electrons to a higher energy level. Notice that the 2p electrons are in different orbitals. According to Hund's rule, we must place one electron into each orbital of the same energy level until all orbitals are half-filled. 2p 2s 1s Oxygen Remember that only two electrons can occupy the same orbital, and that they must be of opposite spin. A different way to represent the ground-state electron configuration is to simply write down the occupied orbitals and to indicate the number of electrons in each orbital. For example, the electron configuration for oxygen is 1s 2 2s 2 2p4 . (b) Nitrogen, with an atomic number of 7, has 7 electrons. Assigning these to energy levels: Nitrogen 2p 2s 1s The more concise way to represent ground-state electron configuration for nitrogen: 1s 2 2s 2 2p3 (c) Sulfur has 16 electrons. 1s 2 2s 2 2p6 3s 2 3p4 Sulfur 2p 2s 1s 3s 3p Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Structure and Bonding 5 1.2 The elements of the periodic table are organized into groups that are based on the number of outer-shell electrons each element has. For example, an element in group 1A has one outershell electron, and an element in group 5A has five outer-shell electrons. To find the number of outer-shell electrons for a given element, use the periodic table to locate its group. (a) Magnesium (group 2A) has two electrons in its outermost shell. (b) Cobalt is a transition metal, which has two electrons in the 4s subshell, plus seven electrons in its 3d subshell. (c) Selenium (group 6A) has six electrons in its outermost shell. 1.3 A solid line represents a bond lying in the plane of the page, a wedged bond represents a bond pointing out of the plane of the page toward the viewer, and a dashed bond represents a bond pointing behind the plane of the page. C Chloroform H Cl Cl Cl 1.4 C C Ethane H H H H H H 1.5 Identify the group of the central element to predict the number of covalent bonds the element can form. (a) Carbon (Group 4A) has four electrons in its valence shell and forms four bonds to achieve the noble-gas configuration of neon. A likely formula is CCl4. Element Group Likely Formula (b) Al 3A AlH3 (c) C 4A CH2Cl2 (d) Si 4A SiF4 (e) N 5A CH3NH2 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 6 Chapter 1 1.6 Start by drawing the electron-dot structure of the molecule. (1) Determine the number of valence, or outer-shell electrons for each atom in the molecule. For chloroform, we know that carbon has four valence electrons, hydrogen has one valence electron, and each chlorine has seven valence electrons. C H Cl total valence electrons 4 x 1 = 4 1 x 1 = 1 7 x 3 = 21 26 . . . . . . : .. .. (2) Next, use two electrons for each single bond. Cl C H .. ..:: Cl Cl (3) Finally, use the remaining electrons to achieve an noble gas configuration for all atoms. For a line-bond structure, replace the electron dots between two atoms with a line. Cl C H .. ..:: Cl Cl .. .. : : .. .. ..:: (a) CHCl3 C Cl Cl Cl H (b) H2S S H 8 valence electrons H (c) CH3NH2 14 valence electrons CH H H N H H : : S .. .. H H (d) CH3Li 8 valence electrons CH H H Li H : : C Li .. .. H H .. .. : : ..:: : .. .. .. .. Molecule Electron-dot structure Line-bond structure .. .. : : .. .. H H CH H .. ..:: N H 1.7 Each of the two carbons has 4 valence electrons. Two electrons are used to form the carbon–carbon bond, and the 6 electrons that remain can form bonds with a maximum of 6 hydrogens. Thus, the formula C2H7 is not possible. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Structure and Bonding 7 1.8 Connect the carbons and add hydrogens so that all carbons are bonded to four different atoms. C C C H H H H H H H H C C C H H H H H H H H sp3 sp3 sp3 Propane The geometry around all carbon atoms is tetrahedral, and all bond angles are approximately 109°. 1.9 C C C C C H C H HHH HHH HHH HHH C CCCCC H H H H H H H H H H H H H H Hexane 1.10 C H H H C C H H H C C C H H H H H H sp3 Propene sp2 sp2 1 23 The C3–H bonds are σ bonds formed by overlap of an sp3 orbital of carbon 3 with an s orbital of hydrogen. The C2–H and C1–H bonds are σ bonds formed by overlap of an sp2 orbital of carbon with an s orbital of hydrogen. The C2–C3 bond is a σ bond formed by overlap of an sp 3 orbital of carbon 3 with an sp2 orbital of carbon 2. There are two C1–C2 bonds. One is a σ bond formed by overlap of an sp 2 orbital of carbon 1 with an sp2 orbital of carbon 2. The other is a π bond formed by overlap of a p orbital of carbon 1 with a p orbital of carbon 2. All four atoms connected to the carbon–carbon double bond lie in the same plane, and all bond angles between these atoms are 120°. The bond angle between hydrogen and the sp3 -hybridized carbon is 109°. 1.11 C H H C C C H H H H All atoms lie in the same plane, and all bond angles are approximately 120°. sp2 sp2 sp2 sp2 1 2 3 4 1,3-Butadiene Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8 Chapter 1 1.12 C C C C C C H H H C O O H O C O C H H H H sp3 .. .. : : .. .. .. : Aspirin. All carbons are sp2 hybridized, with the exception of the indicated carbon. All oxygen atoms have two lone pairs of electrons. 1.13 C H H H CCH C C H H H C H sp3 Propyne 123 sp sp The C3-H bonds are σ bonds formed by overlap of an sp3 orbital of carbon 3 with an s orbital of hydrogen. The C1-H bond is a σ bond formed by overlap of an sp orbital of carbon 1 with an s orbital of hydrogen. The C2-C3 bond is a σ bond formed by overlap of an sp orbital of carbon 2 with an sp3 orbital of carbon 3. There are three C1-C2 bonds. One is a σ bond formed by overlap of an sp orbital of carbon 1 with an sp orbital of carbon 2. The other two bonds are π bonds formed by overlap of two p orbitals of carbon 1 with two p orbitals of carbon 2. The three carbon atoms of propyne lie in a straight line: the bond angle is 180°. The H–C1≡C2 bond angle is also 180°. The bond angle between hydrogen and the sp3 - hybridized carbon is 109°. 1.14 The sp3-hybridized oxygen atom has tetrahedral geometry. (a) C O C H H H H H H .. .. Tetrahedral geometry at nitrogen and carbon. (b) N CH3 H3C H3C .. Like nitrogen, phosphorus has five outer-shell electrons. PH3 has tetrahedral geometry. (c) P H H .. H Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Structure and Bonding 9 The sp3-hybridized sulfur atom has tetrahedral geometry. (d) .. .. S CH2CH2CHCOH NH2 O H3C 1.15 Remember that the end of a line represents a carbon atom with 3 hydrogens, a two-way intersection represents a carbon atom with 2 hydrogens, a three-way intersection represents a carbon with 1 hydrogen and a four-way intersection represents a carbon with no hydrogens. 0 H (b) O HO 3 H 2 H 2 H 2 H 2 H 2 H 2 H 1 H 1 H 1 H 1 H 0 H 0 H 0 H (a) HO HO OH NHCH3 1 H 1 H 1 H 0 H 0 H 0 H 1 H 2 H Adrenaline – C9H13NO3 Estrone – C18H22O2 3 H 1.16 Several possible skeletal structures can satisfy each molecular formula. (a) C5H12 C2H7N NH2 NH (b) (c) C3H6O O O O O OH OH OH OH O (d) C4H9Cl Cl Cl Cl Cl Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 10 Chapter 1 1.17 H2N O OH PABA Visualizing Chemistry 1.18 N H H C N C C C C H H H H H H H C C CH3 H H H H H H C8H17N (a) CC H H H COH N H H H O : : .. .. : NH2 O OH C3H7NO2 (b) 1.19 Citric acid (C6H8O7) contains seven oxygen atoms, each of which has two electron lone pairs. Three of the oxygens form double bonds with carbon. C O O H C H H C C C O O H H H OC O H H O .. .. : : .. .. : : :: : .. .. .. Citric acid Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Structure and Bonding 11 1.20 All carbons are sp2 hybridized, except for the carbon indicated as sp3. The two oxygen atoms and the nitrogen atom have lone pair electrons, as shown. Acetaminophen C C C CC C HH HH O H N C C O H H H H sp3 : .. : : : 1.21 Aspartame C C C C C C C C C O O CH3 N H CC H2N O C C O HO H H H H H H H H H H H Additional Problems Electron Configuration 1.22 Atomic Number of Element Number valence electrons (a) Zinc 30 2 (b) Iodine 53 7 (c) Silicon 14 4 (d) Iron 26 2 ( in 4s subshell), 6 (in 3d subshell) 1.23 Atomic Ground-state Element Number electron configuration (a) Potassium 19 1s 2 2s 2 2p6 3s 2 3p6 4s 1 (b) Arsenic 33 1s 2 2s 2 2p6 3s 2 3p6 4s 2 3d10 4p3 (c) Aluminum 13 1s 2 2s 2 2p6 3s 2 3p1 (d) Germanium 32 1s 2 2s 2 2p6 3s 2 3p6 4s 2 3d10 4p2 Electron-Dot and Line-Bond Structures 1.24 (a) NH2OH (b) AlCl3 (c) CF2Cl2 (d) CH2O Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 12 Chapter 1 1.25 (a) The 4 valence electrons of carbon can form bonds with a maximum of 4 hydrogens. Thus, it is not possible for the compound CH5 to exist. (b) If you try to draw a molecule with the formula C2H6N, you will see that it is impossible for both carbons and nitrogen to have a complete octet of electrons. Therefore, C2H6N is unlikely to exist. (c) A compound with the formula C3H5Br2 doesn't have filled outer shells for all atoms and is thus unlikely to exist. 1.26 : : C C ::: N : .. .. H H H Acetonitrile In the compound acetonitrile, nitrogen has eight electrons in its outer electron shell. Six are used in the carbon-nitrogen triple bond, and two are a nonbonding electron pair. 1.27 : : .. C C H H H Cl Vinyl chloride Vinyl chloride has 18 valence electrons. Eight electrons are used for 4 single bonds, 4 electrons are used in the carbon–carbon double bond, and 6 electrons are in the 3 lone pairs that surround chlorine. 1.28 (a) (b) (c) .. : : : : C O H3C NH2 C O H3C O .. ..: S S CH3 H3C – .. .. .. .. 1.29 In molecular formulas of organic molecules, carbon is listed first, followed by hydrogen. All other elements are listed in alphabetical order. Compound Molecular Formula (a) Aspirin C9H8O4 (b) Vitamin C C6H8O6 (c) Nicotine C10H14N2 (d) Glucose C6H12O6 Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Structure and Bonding 13 1.30 To work a problem of this sort, you must draw all possible structures consistent with the rules of valence. You must systematically consider all possible attachments, including those that have branches, rings and multiple bonds. C C C H H H H H H H H C N H HH H H C COH H H H H H C O C H H H H H H (a) (b) (c) C C C Br H HH HH H H C C C Br H H HH H H H (d) C CH H H O H C CH H H O H C C O H H H H (e) C C C N H HH H H H H HH C CCH HHH H H H N H H C CNC H H HHH HH H H H NC C H H H H H C H H H (f) 1.31 C CH3 H OH H Ethanol 1.32 COH O CCC O O H O H H C CCCOH O O H O H OH H CCC CO O O H O H OH 1.33 C NC C H H H C C H H H H H O C H H H C CCC H H H H H H H H H H (a) (b) (c) C C C CC C HH H H H H H H H H (d) .. .. : Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 14 Chapter 1 1.34 C H H H O– K+ ionic All other bonds are covalent. Hybridization 1.35 The H3C– carbon is sp3 hybridized, and the –CN carbon is sp hybridized. 1.36 CH3CH2CH3 (a) sp3 sp3 sp3 (b) C CH2 H3C H3C sp3 sp3 sp2 sp2 H2C C CH CH (c) sp2 sp2 sp sp CH3 C O OH sp3 sp2 (d) 1.37 Benzene C C C CC C H H H HH H All carbon atoms of benzene are sp2 hybridized, and all bond angles of benzene are 120°. Benzene is a planar molecule. 1.38 sp2 sp2 sp3 Glycine Pyridine Lactic acid 120° 120° 109° N C C C C C H H H H H .. C O H2C NH2 OH C O OHC H3C OHH (a) (b) (c) 1.39 Examples: (a) CH3CH2CH CH2 (b) H2C CH CH CH2 (c) H2C CH C CH Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Structure and Bonding 15 1.40 Procaine C C C C O O C CH2 OH HO HO HO OH H (b) sp3 sp3 sp3 sp2 sp2 sp2 Vitamin C C C C C C C C O O CH2 CH2 NH CH2 CH3 CH2 CH3 H2N H H H H + Cl– sp3 sp3 sp3 sp3 sp2 sp2 sp2 sp2 sp2 sp2 sp3 sp3 (a) sp2 1.41 N O P O O – O – C O H HO H3C Pyridoxal phosphate sp3 sp3 sp sp 3 3 sp3 sp2 sp sp2 2 sp2 sp2 sp2 sp2 The bond angles formed by atoms having sp3 hybridization are